Question

What is the 4th term in the expansion of (y+4x^3)^4?

(y+4x^3) ^4

Answer 1

`256yx^9`

Explanation:

It's `(a+b)^n =[color(white)(o)^nC_0*a^(n-0)*b^0] + [color(white)(o)^nC_1*a^(n-1)*b^1] + [color(white)(o)^nC_2*a^(n-2)*b^2] +[color(white)(o)^nC_3*a^(n-3)*b^3] +.........`

First term `color(white)(ww)` `=>(color(white)(i)^4C_0)(y^4)(4x^3)^0`
Second term `color(white)(i)` `=>(color(white)(i)^4C_1)(y^3)(4x^3)^1`
Third term `color(white)(wu)` `=>(color(white)(i)^4C_2)(y^2)(4x^3)^2`
Fourth term `color(white)(ir)` `=>(color(white)(i)^4C_3)(y^1)(4x^3)^3`

Fourth term `=4*y*64x^9` `color(white)(wwwwww)` `["as " color(white)(i)^4C_3 = 4]`

That is, `256yx^9`

Answer 2

`color(blue)(256yx^9)`

Explanation:

For binomial expansions of the form `(y+x)^n`, with `y` descending, we have:

`sum_(r=0)^(n)((n),(r))(y^(n-r)x^r)`

Where:

`((n),(r))=color(white)(0)^nC_(r)=(n!)/((r!(n-r)!)`

We require the 4th term. The power of `y` in the 4th term will be `1`:

So:

`4-r=1=>r=3`

We use this value of `r`:

`((4),(3))(y^(4-3)(4x^3)^3)`

simplifying:

`((4),(3))(64yx^9)`

`((4),(3))=(4!)/(3!(4-3)!)=(4xx3xx2xx1)/(3xx2xx1xx1)=(4xxcancel(3)xxcancel(2)xxcancel(1))/(cancel(3)xxcancel(2)xxcancel(1)xx1)`

`=4/1=4`

`:.`

The 4th term is therefore:

`4*(64yx^9)`

`color(blue)(256yx^9)`